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Question Number 164162 by Zaynal last updated on 15/Jan/22

Prove the; (tan 𝛂 + ((cos 𝛂)/(1 + sin 𝛂))) sin 𝛂 = 𝛂 ^([Z.A])

$$\mathrm{Prove}\:\mathrm{the}; \\ $$$$\left(\boldsymbol{{tan}}\:\boldsymbol{\alpha}\:+\:\frac{\boldsymbol{{cos}}\:\boldsymbol{\alpha}}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{\alpha}}\right)\:\boldsymbol{{sin}}\:\boldsymbol{\alpha}\:=\:\boldsymbol{\alpha} \\ $$$$\:^{\left[\mathrm{Z}.\mathrm{A}\right]} \\ $$

Commented by som(math1967) last updated on 15/Jan/22

{tan𝛂 +((cos𝛂(1βˆ’sin𝛂))/(1βˆ’sin^2 𝛂))}Γ—sinΞ± ={tan𝛂+((cos𝛂(1βˆ’sin𝛂))/(cos^2 𝛂))}Γ—sin𝛂 ={tan𝛂+(1/(cos𝛂)) βˆ’tan𝛂}Γ—sin𝛂 =sec𝛂×sin𝛂=tan𝛂 so (tan𝛂+((cos𝛂)/(1+sin𝛂)))sin𝛂=tan𝛂 ≠𝛂

$$\left\{\boldsymbol{{tan}\alpha}\:+\frac{\boldsymbol{{cos}\alpha}\left(\mathrm{1}βˆ’\boldsymbol{{sin}\alpha}\right)}{\mathrm{1}βˆ’\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\alpha}}\right\}Γ—{sin}\alpha \\ $$$$=\left\{\boldsymbol{{tan}\alpha}+\frac{\boldsymbol{{cos}\alpha}\left(\mathrm{1}βˆ’\boldsymbol{{sin}\alpha}\right)}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\alpha}}\right\}Γ—\boldsymbol{{sin}\alpha} \\ $$$$=\left\{\boldsymbol{{tan}\alpha}+\frac{\mathrm{1}}{\boldsymbol{{cos}\alpha}}\:βˆ’\boldsymbol{{tan}\alpha}\right\}Γ—\boldsymbol{{sin}\alpha} \\ $$$$=\boldsymbol{{sec}\alpha}Γ—\boldsymbol{{sin}\alpha}=\boldsymbol{{tan}\alpha} \\ $$$$\boldsymbol{{so}}\:\left(\boldsymbol{{tan}\alpha}+\frac{\boldsymbol{{cos}\alpha}}{\mathrm{1}+\boldsymbol{{sin}\alpha}}\right)\boldsymbol{{sin}\alpha}=\boldsymbol{{tan}\alpha} \\ $$$$\:\neq\boldsymbol{\alpha} \\ $$

Commented by Zaynal last updated on 15/Jan/22

thank you sir.., answer is tan apha

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}..,\: \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{tan}\:\mathrm{apha} \\ $$

Commented by cortano1 last updated on 15/Jan/22

Commented by som(math1967) last updated on 15/Jan/22

sorry typo i corrected

$${sorry}\:{typo}\:{i}\:{corrected} \\ $$

Commented by Zaynal last updated on 15/Jan/22

thank you sir, very good.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{very}\:\mathrm{good}. \\ $$

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